Shaded leaf area in a square: the double-counting trick

5 August 20265 min read

Question

A square has side 14 cm. Two quadrants of radius 14 cm are drawn from opposite corners and overlap in a leaf-shaped region. Find the area of the shaded leaf. Use π = 22/7.

Shaded leaf area in a square? Most miss this trick!Watch on YouTube

The shaded leaf area is $112\text{ cm}^2$ . The useful idea is not a new circle formula—it is noticing that the leaf is the part counted twice when two quadrants cover the same square.

The diagram has a square of side 14 cm and two quadrants, each with radius 14 cm, drawn from opposite corners. Their overlap forms a leaf. Students often stare at the curved boundary and search for a special “leaf formula”. None is needed.

The blue and green curves are the two quadrant boundaries. There is one leaf: the single shaded overlap between those curves.

Step 1: calculate the two quadrants

With radius 14 cm and $\pi=\frac{22}{7}$ , each quadrant has area

A_{\text{one quadrant}}=\frac14\pi r^2

A_{\text{one quadrant}}=\frac14\times\frac{22}{7}\times14^2

A_{\text{one quadrant}}=\frac14\times\frac{22}{7}\times196

A_{\text{one quadrant}}=\frac14\times22\times28

A_{\text{one quadrant}}=154\text{ cm}^2

So the two quadrants contain

A_{\text{two quadrants}}=154+154

A_{\text{two quadrants}}=308\text{ cm}^2

Step 2: subtract the part counted once

The square's area is

A_{\text{square}}=\text{side}\times\text{side}

A_{\text{square}}=14\times14

A_{\text{square}}=196\text{ cm}^2

The quadrants together cover that square once, with the leaf included one extra time. In inclusion–exclusion language,

A_{\text{quadrant 1}}+A_{\text{quadrant 2}}=A_{\text{square}}+A_{\text{leaf}}

308=196+A_{\text{leaf}}

A_{\text{leaf}}=308-196

\boxed{A_{\text{leaf}}=112\text{ cm}^2}

A second method: two circular segments

Split the leaf along the square's diagonal. Each half is a 90° sector minus a right triangle whose perpendicular sides are both 14 cm.

A_{\text{right triangle}}=\frac12\times14\times14

A_{\text{right triangle}}=98\text{ cm}^2

A_{\text{one segment}}=A_{\text{quadrant}}-A_{\text{right triangle}}

A_{\text{one segment}}=154-98

A_{\text{one segment}}=56\text{ cm}^2

A_{\text{whole leaf}}=2\times56

A_{\text{whole leaf}}=112\text{ cm}^2

Both methods are correct. The double-counting method is shorter here; the sector-minus-triangle method is more flexible when the two curves do not cover the whole square.

The reusable pattern

If the square side and each quadrant radius are both $a$ , the same reasoning gives

A_{\text{leaf}}=2\left(\frac14\pi a^2\right)-a^2

A_{\text{leaf}}=\frac12\pi a^2-a^2

A_{\text{leaf}}=\left(\frac\pi2-1\right)a^2

Now substitute $\pi=\frac{22}{7}$ . Every simplification is shown below:

A_{\text{leaf}}=\left(\frac{22/7}{2}-1\right)a^2

A_{\text{leaf}}=\left(\frac{22}{14}-1\right)a^2

A_{\text{leaf}}=\left(\frac{22}{14}-\frac{14}{14}\right)a^2

A_{\text{leaf}}=\frac{8}{14}a^2

A_{\text{leaf}}=\frac47a^2

Finally, substitute $a=14$ :

A_{\text{leaf}}=\frac47(14)^2

A_{\text{leaf}}=\frac47\times196

A_{\text{leaf}}=4\times28

\boxed{A_{\text{leaf}}=112\text{ cm}^2}

Three checks that prevent shaded-region errors

Name the target. Decide whether you need a union, an overlap, or the unshaded remainder.
Check the boundary. A quadrant here has radius 14 cm, not diameter 14 cm.
Use size as a warning. An answer larger than the containing square can only be an intermediate double-counted total.

The Areas Related to Circles lessons let you peel composite diagrams into larger-minus-smaller and repeated parts before you calculate, followed by guided shaded-region practice.

Practise shaded regions interactively

Your turn

In the double-counting method, which single region is counted twice when the two quadrant areas are added?

Tell us in the comments on YouTube or Instagram, or email connect@youngmathbrains.com. We read every one.

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